# Can you simplify this further?

• April 23rd 2009, 04:45 AM
Peleus
Can you simplify this further?
Hi all,

Had an exam recently and I've got a nagging feeling about a particular question.

$f(x) = \frac{1-x^2}{1-x}$ and $g(t) = t^2$

Find $f(g(t))$ and simplify your answer.

So that results in

$\frac{1-(t^2)^2}{1-t^2}$

Which simplifies to

$\frac{1-t^4}{1-t^2}$

Is there any further simplification we can do? I can't see anything but I have this nagging feeling I'm missing something. Possibly further factorisation taking place considering 1 is also $1^2$ etc leading to some cancelling out.

Did I miss anything?
• April 23rd 2009, 05:15 AM
HallsofIvy
Yes, of course that simplifies. There is a general formula that $a^n- b^n= (a-b)(a^{n-1}+ a^{n-2}b+ \cdot\cdot\cdot+ ab^{n-2}+ b^{n-1}$. In particular
$1- t^2= (1- t)(1+ t)$
and
$1- t^4= (1- t)(1+ t+ t^2+ t^3)$
• April 23rd 2009, 05:22 AM
Peleus
Quote:

Originally Posted by HallsofIvy
Yes, of course that simplifies. There is a general formula that $a^n- b^n= (a-b)(a^{n-1}+ a^{n-2}b+ \cdot\cdot\cdot+ ab^{n-2}+ b^{n-1}$. In particular
$1- t^2= (1- t)(1+ t)$
and
$1- t^4= (1- t)(1+ t+ t^2+ t^3)$

True, however I personally wouldn't consider

$\frac{1+ t+ t^2+ t^3}{1+ t}$

which that cancels down as a "simpler" answer than the original though, would you?
• April 23rd 2009, 05:43 AM
mr fantastic
Quote:

Originally Posted by Peleus
Hi all,

Had an exam recently and I've got a nagging feeling about a particular question.

$f(x) = \frac{1-x^2}{1-x}$ and $g(t) = t^2$

Find $f(g(t))$ and simplify your answer.

So that results in

$\frac{1-(t^2)^2}{1-t^2}$

Which simplifies to

$\frac{1-t^4}{1-t^2}$

Is there any further simplification we can do? I can't see anything but I have this nagging feeling I'm missing something. Possibly further factorisation taking place considering 1 is also $1^2$ etc leading to some cancelling out.

Did I miss anything?

Yes.

$\frac{1-t^4}{1-t^2} = \frac{(1 - t^2)(1 + t^2)}{1 - t^2} = 1 + t^2$, $t \neq \pm 1$.
• April 23rd 2009, 06:45 AM
Peleus
Bother, knew I missed something.

I originally had it as

$= \frac{(1-t^2)^2}{1-t^2}$

$= \frac{(1-t^2)(1-t^2)}{1-t^2}$

$= (1-t^2)$

But realised I wasn't correct because it was never squared, it was $1-(g(t))^2$ so I scribbled it out and didn't spot the other way.